NETWORK THEOREMS (Thevenin's Theorem)

THEVENIN’S THEOREM

 Thevenin’s theorem states that, any linear one-port network can be “replaced with” a single voltage source in series with a single resistor. The voltage source is called the Thevenin’s equivalent voltage, and the resistor is called the Thevenin’s equivalent resistance. The voltage source and series resistor must behave identically to the actual network it is replacing.

The theorem indicates that any linear circuit can be simplified, no matter how complex, to an equivalent circuit with just a single voltage source and series resistance connected to a load.

When we say “linear” circuit we mean circuits with underlying equations with no exponents or roots, this is true of circuits with passive components such as resistors, inductors and capacitors. See:PASSIVE COMPONENTS

However, we may have circuits containing components that their opposition to current flow changes with voltage and/or current (e.g. semiconductor components), circuits containing these types of components are referred to as non-linear circuits.

The advantage of this theorem is that, no matter any change in the value of R_L, we can easily determine the current that flows through it with the Thevenin’s equivalent circuit. Let’s work through some example to understand how it works…

The steps involved in determining the Thevenin’s equivalent circuit for a network:

Find the Thevenin’s voltage and resistance, E_th and R_th respectively for the circuit in figure 1 above.

1. Remove the load resistor
2. Calculate the Open circuit voltage, E_th.
3. Open current source and short voltage source
4. Calculate the open circuit resistance, R_th.
5. Redraw the circuit with E_th and R_th.
6. Find current flowing through load by using the Ohm’s law I_T = E_th/(R_th+R_L)

Example 1:

Find the Thevenin’s equivalent circuit for the network in figure 2, with the load resistance R_L

Figure 1

Step 1: Removing R_L

Here, we can see that E_th is equal to voltage across R₂, V_R2.

Step 2:

 (See: Voltage divider rule).

Step 3:

To determine the relationship between the resistance R₁ and R₂ (i.e. Series or Parallel), we used a means borrowed from "BOYLESTA; “Introductory Circuit Analysis”, TENTH EDITION".  Imagine the trickle current from the ohmmeter, it splits at the node through R₁ and R₂, this depicts the two resistors R₁ and R₂ parallel combination.

There fore;

Step 4:

Step 5:

Step 6:

Now, the current flowing through the load R_L can be deduced by the relation;

For;

R_L = 1Ω, I_L =6/3 = 2A 
R_L = 10Ω,     I_L=0.5A

R_L = 100Ω,     I_L=0.059A

Example 2:

Find the thevenin’s equivalent circuit and the current flowing through the 7Ω resistor in the circuit bellow.

Solution:

Step 1:

E_th is equal to the voltage across R₁, V_R1.

            V_R1=IR₁

From the circuit, current can not pass through R₂ (current does not flow through open circuit), thus, the whole of the 12A current form the source pass through, R₁, therefore;

Step 2:

Step 3:

From the circuit, R₂ is in series connection with R₁||Short circuit (i.e. having zero resistance). The resultant resistance of the parallel combination of R1 and Short circuit is given by;

Step 4:

Therefore;

Step 5:

Step 6:

Current I_L through the 7Ω resistor is thus given by;

Example 3:

The circuit shown in figure 3b is the Thevenin equivalent circuit of that in figure 3a. Find the value of the thevenin’s voltage E_th and Thevenin resistance, R_th.

Solution:

Step 1:

 E_th is equal to the voltage across R₃.  Because we have two sources in the circuit, we will use Superposition rule (See: Superposition Theorem) to determine the voltage across R₃ the thus, E_th.

Now, let remove the 2A current source first;

As no current flow through the open circuits, the current from the voltage source flows as shown by the arrows, thus depict resistors R₁ and R₂ in series connection with the voltage source, E’_th =V_R3 =V_R1.

(See:Voltage divider rule)

Removing the 2V source;

The R₁ and R₃ are in parallel, R₂ and R₃ are in series connection. Eth is the voltage across R₁ and R₃. To determine Eth, we may calculate the voltage across R₁ or that across R₃;

Voltage across R₁ = I₁R₁ (I₁ being the current across R₁)

Step 2:

Step 3:

Using our method as discussed in examples above, the trickle current from the Ohmeter splits between R₁ and R₃, meaning, they are in parallel with each other and R₁ and R₂ are in series connection (Current does not flow to the open circuit end).

We can determine Rth which is the resultant resistance looking-in by;

The  Thevenin’s equivalent circuit is thus as bellow:

See Also:


Voltage divider rule
Current divider rule
Superposition Theorem

Hi readers I'll add some more examples latter ...